All categories

4 years ago

PLEASE HELP ASAP!!!!!! What should each experiment only have one of? variable or constant

Physics

2 answers:

d1i1m1o1n [39]4 years ago

4 0

Variable - meaning one element, feature or factor able to change

slega [8]4 years ago

3 0

Answer:

the answer should be a constant

hope this helps!!

You might be interested in

Who ever gets it I will give Brainly and 20 points

Likurg_2 [28]

What are the options for A,B,C,D? There’s arrows

5 0

3 years ago

Read 2 more answers

Two objects have charges 2.0 C and 1.0 C. If the objects are placed 2 meters apart, what is the magnitude of the force that the

kondor19780726 [428]

Answer:

$4.5\cdot 10^9 N$

Explanation:

The electric force between two charged objects is given by:

$F=k\frac{q_1 q_2}{r^2}$

where:

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is their separation

In this problem:

q1 = 2.0 C

q2 = 1.0 C

r = 2 m

So, the electric force is

$F=(9\cdot 10^9 Nm^2C^{-2})\frac{(2.0 C)(1.0 C)}{(2 m)^2}=4.5\cdot 10^9 N$

4 0

4 years ago

You can make a solute dissolve more quickly in a solvent by

Gekata [30.6K]

Answer:Stirring.

Explanation:Stirring a solute into a solvent speeds up the rate of dissolving because it helps distribute the solute particles throughout the solvent. For example, when you add sugar to iced tea and then stir the tea, the sugar will dissolve faster.

3 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

When the north pole of a magnet is moved towards the center of a loop of wire containing a galvanometer, the needle of the galva

andriy [413]

Answer:

Moving the magnet away from the center of the loop with its south pole facing the center of the loop.

Explanation:

Electromagnetic induction is due to a rapidly changing magnetic field, or loop area. The poles of the magnet induce current in the loop but in the opposite direction, depending on the direction of their relative motion. An approaching north pole will induce an anticlockwise current in the loop, while an approaching south pole will do the reverse. To get the galvanometer to flicker in the same direction as of that when the north pole was approaching, we move the magnet away from the center of the loop with its south pole facing the center of the loop.

6 0

3 years ago